Determine the equation of a hyperbola with foci at $(3,7)$ and $(3,−1)$ and with eccentricity $e=2$.
If someone could check my answer that would be great.
By looking at the foci it is easy to deduce that the equation of the hyperbola will be of the form $$\frac{x^2}{b^2}-\frac{y^2}{a^2}=1~~(a>b)$$
We also know that the center will be at $(3,3)$ and so $c=4$ also for a hyperbola $e=\frac{c}{a}=\frac{4}{a}=2 \iff a=2 \implies a^2=4~$ and also $b^2=a^2+c^2 \iff b^2=20$ thus we have the equation as $$\frac{x^2}{20}-\frac{y^2}{4}=1$$
Thanks.
If this is wrong please tell me how so.
As the foci lies in $x=3$ which is the major axis
The midpoint of the foci is centre $(3,3)$
Now the coordinate of the foci $(\alpha,ae\pm\beta)$ where $2a=$ major axis, $(\alpha,\beta)$ is the center and $e=$eccentricity
Finally $b^2=a^2(e^2-1)$ where $2b=$minor axis
Hope this should be sufficient